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A081090
a(n)^2 + 1 = A081089(n), where A081089(n) = A081088(n+1)/A081088(n); involves the partial quotients of a series of continued fractions that sum to unity.
4
0, 2, 5, 52, 13525, 9512132552, 1223751003414213892335125, 14245051228051734585272181044575005954679284643762013257552
OFFSET
1,2
COMMENTS
log(a(n+1))/log(a(n)) -> 1+sqrt(2). The 8th term has 59 digits, while the 9th term has 141 digits.
sum(n>0, a(n)/a(n+1) ) = 1/2; the ratio of the terms a(n+1)/a(n), for n>1, form the convergents of the continued fraction series described by A081088; thus a(n+1) = A081088(n)*a(n) + a(n-1), for n>1. - Paul D. Hanna, Mar 05 2003
FORMULA
a(n) = a(n-2)*(a(n-1)^2 + 1) for n>3, with a(1)=0, a(2)=2, a(3)=5. Also, a(n)*a(n-1) = A081088(n) for n>2, a(n) = a(n-2)*A081089(n-1) for n>2.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Hans Havermann and Paul D. Hanna, Mar 05 2003
STATUS
approved